Subtopic: Steering and hydraulic control systems used in tractors
Question Statement:
A double acting hydraulic cylinder has a piston diameter of 40 mm and the rod diameter equal to one-half the piston diameter. For a constant pressure of 4 MPa, the difference in load carrying capacity between extension and retraction is
Options:
(A) 0 kN
(B) 1.26 kN
(C) 3.77 kN
(D) 6.29 kN
Official Answer Key:
B
Step-by-Step Solution & Explanation
Piston area A_p = π/4×0.04² = 1.2566×10⁻³ m². Rod area A_r = π/4×0.02² = 3.1416×10⁻⁴ m². Extension force F_ext = p×A_p = 4×10⁶×1.2566×10⁻³ = 5026.5 N. Retraction force F_ret = p×(A_p−A_r) = 4×10⁶×9.425×10⁻⁴ = 3769.9 N. Difference = 5026.5−3769.9 = 1256.6 N ≈ 1.26 kN.
Practice This in Real CBT Exam Interface
Test yourself with the official TCS-iON palette, virtual calculator, and immediate performance analytics.